Question: Evaluate $~~\int \tan^{-1}(2x)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\tan^{-1}(2x)-\dfrac14\ln(4x^2+1)+C$ (Choice B) B $x\tan^{-1}(2x)-\dfrac14\ln(4x^2+1)+C$ (Choice C) C $x\tan^{-1}(2x)-\dfrac1{16}\ln(4x^2+1)+C$ (Choice D) D $\tan^{-1}(2x)-\dfrac1{16}\ln(4x^2+1)+C$
Answer: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \tan^{-1}(2x)~$ and $~dv=dx\,$. Then $~du = \dfrac{2}{4x^2+1}\,dx~$ and $~v = \int dx = x\,$. Integration by parts gives $ \int \tan^{-1}(2x)\,dx = x\tan^{-1}(2x)-\int\dfrac{2x}{4x^2+1}\,dx$ $ ~\,=x\tan^{-1}(2x)-\dfrac14\ln(4x^2+1)+C\,$.